Amplitude of the complex number i sin ((π)/19 ) is

Correct answer: (C) (π)/2. Let z=i sin (π)/19 or z=0+i sin (π)/19 Let amplitude of z be θ , then sin θ=fracsin (π)/19√0^2+ (sin (π)/19 )^2 [∵ sin θ=fracb√a^2+b^2 ] or sin θ=(sin (π)/19)/(sin (π)/19)=1 ⇒ θ=(π)/2

Q72 / 180 · COMEDK 2015 · Mathematics

Amplitude of the complex number $i \sin \left(\frac{\pi}{19}\right)$ is

A
$\frac{\pi}{19}$
B
$-\frac{\pi}{19}$
Correct answer
C
$\frac{\pi}{2}$
D
$\frac{\pi}{2}-\frac{\pi}{19}$

Solution

Let $z=i \sin \frac{\pi}{19}$ or $\quad z=0+i \sin \frac{\pi}{19}$ Let amplitude of $z$ be $\theta$, then $$ \sin \theta=\frac{\sin \frac{\pi}{19}}{\sqrt{0^{2}+\left(\sin \frac{\pi}{19}\right)^{2}}}\left[\because \sin \theta=\frac{b}{\sqrt{a^{2}+b^{2}}}\right] $$ or $\sin \theta=\frac{\sin \frac{\pi}{19}}{\sin \frac{\pi}{19}}=1 \Rightarrow \theta=\frac{\pi}{2}$

Hence, the correct answer is option: 'C'.

Amplitude of the complex number $i \sin \left(\frac{\pi}{19}\right)$ is

Solution

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